Answer:
5 mg, 
Explanation:
First of all, let's rewrite the mass in grams using scientific notation.
we have:
m = 0.005 g
To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is

If we want to convert into milligrams, we must remind that
1 g = 1000 mg
So we can use the proportion

and we find

Answer:
y = 54.9 m
Explanation:
For this exercise we can use the relationship between the work of the friction force and mechanical energy.
Let's look for work
W = -fr d
The negative sign is because Lafourcade rubs always opposes the movement
On the inclined part, of Newton's second law
Y Axis
N - W cos θ = 0
The equation for the force of friction is
fr = μ N
fr = μ mg cos θ
We replace at work
W = - μ m g cos θ d
Mechanical energy in the lower part of the embankment
Em₀ = K = ½ m v²
The mechanical energy in the highest part, where it stopped
= U = m g y
W = ΔEm =
- Em₀
- μ m g d cos θ = m g y - ½ m v²
Distance d and height (y) are related by trigonometry
sin θ = y / d
y = d sin θ
- μ m g d cos θ = m g d sin θ - ½ m v²
We calculate the distance traveled
d (g syn θ + μ g cos θ) = ½ v²
d = v²/2 g (sintea + myy cos tee)
d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)
d = 1555.85 /7.8145
d = 199.1 m
Let's use trigonometry to find the height
sin 16 = y / d
y = d sin 16
y = 199.1 sin 16
y = 54.9 m
The crate only moves horizontally, so its net vertical force is 0. The only forces acting in the vertical direction are the crate's weight (pointing downward) and the normal force of the surface on the crate (pointing upward). By Newton's second law, we have
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0 → <em>n</em> = <em>mg</em> = 1876 N
where <em>n</em> is the magnitude of the normal force.
In the horizontal direction, the crate is moving at a constant speed and thus with no acceleration, so it's completely in equilibrium and the net horizontal force is also 0. The only forces acting on it in this direction are the 747 N push (pointing in the direction of the crate's motion) and the kinetic friction opposing it (pointing in the opposite direction). By Newton's second law,
∑ <em>F</em> (horizontal) = 747 N - <em>f</em> = 0 → <em>f</em> = 747 N
The frictional force is proportional to the normal force by a factor of the coefficient of kinetic friction, <em>µ</em>, such that
<em>f</em> = <em>µn</em> → <em>µ</em> = <em>f</em> / <em>n</em> = (747 N) / (1876 N) ≈ 0.398188 ≈ 0.40
Answer:
vₓ = 20 m/s, v_{y} = -15 m / s
Explanation:
This is a conservation of moment problem, since it is a vector quantity we can work each axis independently
The system is formed by the two drones, so the forces during the crash are internal and the moment is conserved
X axis
Initial moment. Before the crash
p₀ = m₁ v₀ₓ + m₂ v₀ₓ
Final moment. After the crash
p_{fx} = (m₁ + m₂) vₓ
p₀ₓ =
m₁ v₀ₓ + m₂ v₀ₓ = (m₁ + m₂) vₓ
vₓ = (m₁ + m₂) v₀ₓ / (m₁ + m₂)
vₓ = v₀ₓ = 20 m/s
Y Axis
Initial
p_{oy} = m₁ v_{oy}
Final
p_{fy} = (m₁ + m₂) v_{y}
p_{oy} = p_{fy}
the drom rises and when it falls it has the same speed because there is no friction v_{oy} = -60 m/s
m₁
= (m₁ + m₂) v_{y}
v_{y} = m₁ / (m₁ + m₂) v_{oy}
v_{y} = 1/4 60
v_{y} = -15 m / s
Vertical speed is down